1
$\begingroup$

I'm self studying analytic number theory and I wonder how the proof goes for this problem: If $f (x) $ satisfies $f (x)=x^2+O (x) $, and $f $ is differentiable with nondecreasing derivative $f'(x) $ for sufficiently large $x $, then $f'(x)=2x+O (\sqrt{x}) $.

I have a proof but I did not use the fact that $f'$ is nondecreasing so I bet that my proof is wrong. Any help will be greatly appreciated.

It can be shown that for sufficiently large x, $f'(x)=2x+O (1) $. Basically, I just used the hypothesis and the limit definition of the derivative. So now, by dividing by $\sqrt {x} $, we obtain $\dfrac {f'(x)}{\sqrt{x}}=2\sqrt {x}+O \left(\dfrac {1}{\sqrt {x}}\right)=2\sqrt {x}+O (1) $. So multiplying by $\sqrt {x} $ we get the desired result. But I'm pretty sure that there is a flaw in my proof.

$\endgroup$
  • $\begingroup$ If you show us your proof, we might tell you where it goes wrong (if it does). $\endgroup$ – Yves Daoust Oct 17 '17 at 7:04
2
$\begingroup$

It holds $$x^m = \mathcal{O}(x^n),\qquad m≤n, \qquad x→∞,$$

because of $$\frac{x^m}{x^n} = x^{m-n} →\begin{cases}0 &m≤n\\1 & m=n \\ ∞ &m≥n\end{cases}\quad \text{for } x→∞.$$

Therefore if $g(x)=\mathcal{O}(1)=\mathcal{O}(x^0)$ it follows $g(x)=\mathcal{O}(x^{\frac{1}{2}})$ for $x→∞$.

This should help you proof the desired result, with what you already have done so far.

The information $f'$ to be nondecreasing is an information, which is not necessary for the proof. IMO that can be best seen by the following argument, instead of a rigorous proof.

If $f(x)=x^2+\mathcal{O}(x)$, then it is $f'(x)=x+\mathcal{O}(1)$. Let $f'(x)=h(x)+g(x)$, with $h(x)=x$ and $g$ being that function that behaves like $\mathcal{O}(1)$.

This means $$\frac{g(x)}{1}→c∈ℝ,\qquad x→∞.$$

And that means, that the governing part of $f'$ is $h(x)=x$, because $g$ can not grow as fast as $h$ does. So for sufficiently large $x$ the behaviour of $f'$ will be like $h$, and $h$ is nondecreasing.


Edit:
I somehow missed the last equation of your proof. What you did is correct, since you used what I stated in this answer, in your last "="-step.

$\endgroup$
  • $\begingroup$ Woah thanks for enlightening me about the information about $f'$. I just realized that in studying these estimates, I really have to understand what is going on. Thank you very much! $\endgroup$ – Habagat Maliksi Oct 17 '17 at 8:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.